02/24/10 Lecture

Announcements

• Corrections to Monday’s Powerpoint Lecture Slides Posted (Slides 3, 9 and 15)

• I have posted last semester’s Exam 1 plus the key

• Exam 1 – Topics– Ch. 1, 3, and 4 (all topics covered in

reading/lectures)– Ch. 6 (Sections 1 through 4 expected)

Today’s Lecture

• Chapter 6 Material– Thermodynamics

• Definitions• Determination of signs of ΔS, ΔH, ΔG• Relationship between ΔG, Q and K• Le Chatelier’s Principle

– Solubility Product Problems

Thermodynamics

Some Definitions:H = EnthalpyS = EntropyG = Gibbs Free EnergyT = Absolute temperature

Thermodynamics

1. ΔH is related to heat of reaction- if a reaction produces heat, ΔH < 0 and

reaction is “exothermic”- a reaction that requires heat has ΔH >

0 and is endothermic

2. ΔS is related to disorder of system- If the final system is “more random”

than initial system, ΔS > 0

Thermodynamics

Entropy Examples: (Is ΔS > or < 0?)H2O(l) ↔ H2O(g)

H2O(s) ↔ H2O(l)

NaCl(s) ↔ Na+ + Cl-

2H2(g) + O2(g) ↔ 2H2O(g)

N2(g) + O2(g) ↔ 2NO(g)

ΔS > 0

ΔS > 0ΔS > 0

ΔS < 0

ΔS > 0

Thermodynamics

ΔG = Change in Gibbs free energyThis tells us if a process is spontaneous

(expected to happen) or non-spontaneous

ΔG < 0 process is spontaneous (favored)ΔG = ΔH - TΔS (T is absolute temperature)

processes that are exothermic (Δ H < 0) and increase disorder (Δ S > 0) are favored at all Tprocesses that have Δ H > 0 and Δ S > 0 are favored at high T

Example question

The reaction N2(g) + O2 (g) ↔ 2NO(g) has a positive H. Under what conditions is this process spontaneous?- all temperatures- low temperatures- high temperatures- never

Thermodynamics

ΔG and EquilibriumΔG = ΔG° + RTlnQ Q = Reaction

Quotient(for A ↔ B, Q = [B]/[A])At equilibrium, ΔG = 0 and Q = K

ΔG° = -RTlnK

Thermodynamics

• Example Question:The ΔG° for the reactionCa2+ + 2OH- => Ca(OH)2(s) is -52 kJ/mol

Determine K at T = 20°C for Ca(OH)2(s) => Ca2+ + 2OH-

Section 6 – 2: Le Châtelier’s Principle

• The position of a chemical equilibrium always shifts in a direction that tends to relieve the effect of an applied stress

• Types of stresses:– Addition/removal of reactant/product– Change in volume (gases) or dilution (aqueous solutions)– Changes in temperatures

• Can take intuitive or mathematical approaches to solving problems

Le Châtelier’s Principle

Intuitive Method– Addition to one side

results in switch to other side

– Example:

Mathematical Method

AgCl(s) ↔ Ag+ + Cl-

Additional Ag+ results in more AgCl

QRTGG ln

K

QRTG ln

When Q>K, ΔG>0 (toward reactants)

When Q<K, ΔG<0 (toward products)

Example: Q = [Ag+][Cl-]

As Ag+ increases, Q>K

Le Châtelier’s Principle

Stress Number 1 Reactant/Products:

Addition of reactant: shifts toward product

Removal of reactant: shifts toward reactant

Addition of product: shifts toward reactant

Removal of product: shifts toward product

Le Châtelier’s Principle

Stress Number 1 Example:CaCO3(s) + 2HC2H3O2(aq) ↔ Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)

1. Add HC2H3O2(aq)

2. Remove CO2(g)

3. Add Ca(C2H3O2)2(aq)

4. Add CaCO3(s) No effect because (s)

Le Châtelier’s Principle

Stess Number Two: Dilution

Side with more moles is favored at lower concentrations

Example: HNO2(aq) ↔ H+ + NO2-

If solution is diluted, reaction goes to products

If diluted to 2X the volume:

2

2

HNONOH

K

2

2

21

21

21

HNO

NOHQ

KQ2

1

So Q<K, products favored

Le Châtelier’s PrincipleStess Number Two: Dilution – Molecular Scale View

H+ NO2-

Concentrated Solution

H+ NO2-

H+

NO2-

H+ NO2-

H+ NO2-

Diluted Solution – dissociation allows ions to fill more space

H+ NO2-

H+ NO2-

H+

NO2-

H+ NO2-

H+ NO2-

Le Châtelier’s Principle

Stress Number 3: TemperatureIf ΔH>0, as T increases, products favoredIf ΔH<0, as T increases, reactants favoredEasiest to remember by considering heat

a reactant or productExample:

OH- + H+ ↔ H2O(l) + heatIncrease in T

Solubility Product Problems

Importance:- gravimetric analysis

- chemical separations (e.g. selective removal of Mg2+ or Ca2+ to determine single ion in water hardness titration)

Solubility Product Problems - Solubility in Water

Example: solubility of Mg(OH)2 in waterSolubility defined as mol Mg(OH)2 dissolved/L

sol’n or g Mg(OH)2 dissolved/L sol’n or other units

Use ICE approach:Mg(OH)2(s) ↔ Mg2+ + 2OH-

Initial 0 0Change +x +2xEquilibrium x 2x

Solubility Product Problems

- Solubility of Mg(OH)2 in water

Equilibrium Equation: Ksp = [Mg2+][OH-]2

Ksp = 7.1 x 10-12 = x(2x)2 = 4x3 (see Appendix F for Ksp)

x = (7.1 x 10-12/4)1/3 = 1.2 x 10-4 M

Solubility = moles Mg(OH)2 dissolved = x

Solubility = 1.2 x 10-4 MConc. [OH-] = 2x = 2.4 x 10-4 M

Solubility Product Problems

- Solubility of Mg(OH)2 in Common Ion

If we dissolve Mg(OH)2 in a common ion (OH- or Mg2+), from Le Châtelier’s principle, we know the solubility will be reduced

Example 1) What is the solubility of Mg(OH)2 in a pH = 11.0 buffer?

Ksp = 7.1 x 10-12

Solubility Product Problems

- Solubility of Mg(OH)2 in Common Ion

Example 2) Solubility of Mg(OH)2 in 5.0 x 10-3 M MgCl2.

Solubility Product Problems Precipitation Problems

What occurs if we mix 50 mL of 0.020 M BaCl2 with 50 mL of 3.0 x 10-4 M (NH4)2SO4?

Does any solid form from the mixing of ions?

What are the concentrations of ions remaining?

Precipitations Used for Separations

Example: If we wanted to know the concentrations of Ca2+ and Mg2+ in a water sample. EDTA titration gives [Ca2+] + [Mg2+]. However, if we could selectively remove Ca2+ or Mg2+ (e.g. through titration) and retitrate, we could determine the concentrations of each ion.

Determine if it is possible to remove 99% of Mg2+ through precipitation as Mg(OH)2 without precipitating out any Ca(OH)2 if a tap water solution initially has 1.0 x 10-3 M Mg2+ and 1.0 x 10-3 M Ca2+.